You are viewing an old version of this page. View the current version.
Compare with Current
View Page History
« Previous
Version 6
Next »
Unknown macro: {table}
Unknown macro: {tr}
Unknown macro: {td}
Error formatting macro: live-template: java.lang.NullPointerException
Unknown macro: {td}
null
Part A
A person pushes a box of mass 15 kg along a floor by applying a perfectly horizontal force F. The box accelerates horizontally at a rate of 2.0 m/s2. Assuming the coefficient of kinetic friction between the box and the ground is 0.45, what is the magnitude of F?
Solution
System: Box as point particle.
Interactions: External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).
Model: Point Particle Dynamics.
Approach:
Diagrammatic Representation
The free body diagram for this situation is:
![](/confluence/download/attachments/32278191/pushingboxfrfbd1.png?version=1&modificationDate=1229885157000&api=v2)
Mathematical Representation
With this free body diagram, Newton's 2nd Law can be written:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ \sum F_
Unknown macro: {x}
= F - F_
Unknown macro: {f}
= ma_
]
[ \sum F_
Unknown macro: {y} = N - mg = ma_
= 0 ]\end
where we have assumed that the y acceleration is zero because the box is sliding along a horizontal floor, not moving upward or downward. This realization is important, because we know Ff = μN. Thus, because the y acceleration is zero, we can solve Newton's 2nd Law in the y direction to yield:
Unknown macro: {latex}
\begin
Unknown macro: {large} [ N = mg]\end
so that:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ F = ma_
Unknown macro: {x}
+F_
Unknown macro: {f}
= ma_
+ \mu_
Unknown macro: {k}
N = ma_
Unknown macro: {x}
+ \mu_
mg = \mbox
Unknown macro: {96 N}
] \end
Part B
A person pushes a box of mass 15 kg along a floor by applying a perfectly horizontal force F. The box moves horizontally at a constant speed of 2.0 m/s in the direction of the person's applied force. Assuming the coefficient of kinetic friction between the box and the ground is 0.45, what is the magnitude of F?
System, Interactions and Model: As in Part A.
Approach:
Just as above, Newton's 2nd Law can be written:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ \sum F_
Unknown macro: {x}
= F - F_
Unknown macro: {f}
= ma_
]
[ \sum F_
Unknown macro: {y}
= N - mg = 0] \end
This time, however, the x acceleration is also zero, since the box maintains a constant speed. This implies:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ F = ma_
Unknown macro: {x}
+ F_
Unknown macro: {f} = F_
= \mu_
Unknown macro: {k}
mg = \mbox
Unknown macro: {66 N}
] \end
Error formatting macro: live-template: java.lang.NullPointerException